Question: Find the slope and y-intercept of the line that is ${\text{perpendicular}}$ to $\enspace {y = x }\enspace$ and passes through the point ${(2, -5)}$. ${1}$ ${2}$ ${3}$ ${4}$ ${5}$ ${6}$ ${7}$ ${8}$ ${9}$ ${\llap{-}2}$ ${\llap{-}3}$ ${\llap{-}4}$ ${\llap{-}5}$ ${\llap{-}6}$ ${\llap{-}7}$ ${\llap{-}8}$ ${\llap{-}9}$ ${1}$ ${2}$ ${3}$ ${4}$ ${5}$ ${6}$ ${7}$ ${8}$ ${9}$ ${\llap{-}2}$ ${\llap{-}3}$ ${\llap{-}4}$ ${\llap{-}5}$ ${\llap{-}6}$ ${\llap{-}7}$ ${\llap{-}8}$ ${\llap{-}9}$
Answer: Lines are considered perpendicular if their slopes are negative reciprocals of each other. The slope of the blue line is ${1}$ , and its negative reciprocal is ${-1}$ Thus, the equation of our perpendicular line will be of the form $\enspace {y = -x + b}\enspace$ We can plug our point, $(2, -5)$ , into this equation to solve for ${b}$ , the y-intercept. $-5 = {-}(2) + {b}$ $-5 = -2 + {b}$ $-5 + 2 = {b} = -3$ The equation of the perpendicular line is $\enspace {y = -x - 3}\enspace$. ${m = -1, \enspace b = -3}$